Answer
$\dfrac{k\sqrt{k}+3k}{k-9}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Rationalize the given expression, $
\dfrac{k}{\sqrt{k}-3}
,$ by multiplying the numerator and the denominator by the conjugate of the denominator.
$\bf{\text{Solution Details:}}$
By reversing the operator between the terms of the denominator, $
\sqrt{k}-3
,$ the conjugate is $
\sqrt{k}+3
.$ Multiplying both the numerator and the denominator by the conjugate results to
\begin{array}{l}\require{cancel}
\dfrac{k}{\sqrt{k}-3}\cdot\dfrac{\sqrt{k}+3}{\sqrt{k}+3}
\\\\=
\dfrac{k(\sqrt{k}+3)}{(\sqrt{k}-3)(\sqrt{k}+3)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent\begin{array}{l}\require{cancel}
\dfrac{k(\sqrt{k}+3)}{(\sqrt{k})^2-(3)^2}
\\\\=
\dfrac{k(\sqrt{k}+3)}{k-9}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{k\sqrt{k}+3k}{k-9}
.\end{array}
